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Kryger [21]
3 years ago
9

A wire with radius 23 cm has a current of 7 A which is distributed uniformly through its cross sectional area. If you were to us

e ampere’s law to calculate the magnetic field at a distance of 2 cm from the center of the wire, what would be the current enclosed?
Physics
1 answer:
Rina8888 [55]3 years ago
6 0

Answer:

The magnetic induction of the magnetic field is  0.0005293 mT

Explanation:

Data given

I = 7 A = the total current in the wire

r = 23 cm = the radius of the wire = 0.23 meter

r' = 2cm = the measurement point, which should be inside the wire = 0.02 meter

Let's consider the current density is constant in the wire, ⇒  the current enclosed is a function of the enclosed area

I(enclosed) = Jπ r ²

we can  consider the current density  as the total current over the whole area:

I(enclosed) = I / (πr ²)  * πr' ²

I(enclosed) = (I* r'²)/ (r ²)  

with I =  total current in the wire = 7A

With r = the radius of wire = 0.23 meter

with r' = the distance of point from the center of wire  0.02 meter

We plug this into ampere's law:

∮ *B *dl =μ 0  * (I* r'²)/ (r ²)  

with B = Magnetic flux density (in Tesla) or magnetic induction

with dl = an infinitesimal element (a differential) of the curve C

with µ0 = the magnectic constant =  4π*10^−7 H/m

We can simplify this, by using an Amperian loop can write this as:

B *( 2 π r') =  μ 0  * (I* r'²)/ (r ²)  

Because the circumference of a circle is  2 π r , when we integrate over length at a distance  r ′  from the center of wire whose crossection is a circle we get  2 π r ′

When we isolate B, we get:

B = µo *(Ir'/2 π r ²)

B =  4π*10^−7 * ((7*0.02)/2*π*0.23²)

B =5.293 *10 ^-7 T  = 0.0005293 mT

The magnetic induction of the magnetic field is  0.0005293 mT

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Explanation:

Given

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v is speed of the wave

f is frequency of the wave

λ is wavelength of the wave

therefore frequency of sound wave is given by

f = v/λ

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Hence the frequency of sound wave created by trumpet is 437.5Hz

7 0
3 years ago
An artificial satellite is in a circular orbit around a planet of radius r= 2.05 x103 km at a distance d 310.0 km from the plane
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\rho = 12580.7 kg/m^3

Explanation:

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So here we will have

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here we have

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here we have

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Now the density is given as

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\rho = \frac{4.54 \times 10^{23}}{\frac{4}[3}\pi(2.05 \times 10^6)^3}

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1  greater distances fallen in successive seconds

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Let's look for the position for successive times

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Blood is a homogenous mixture

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